Using code (not the Interface builder) I need to create an NSButton
that looks like an image. Specifically I开发者_Python百科 want to use NSImageNameStopProgressFreestandingTemplate
and I need it not to look like button but to look like the image. This means:
Thanks.
I know this response is a bit late, but you could try this, given thisButton
:
[thisButton setImage:[NSImage imageNamed:NSImageNameStopProgressFreestandingTemplate]];
[thisButton setImagePosition:NSImageOnly];
[thisButton setBordered:NO];
That last line is the key bit: removing the button border effectively strips it of its bezel, leaving only the image to click on. (BTW, I haven't tried the above code specifically, so you may need to throw in a couple of other tweaks, such as setting the imageScaling
or buttonType
, to get it to work best.)
One final note: If you're using a template image (as you said you would), Cocoa will automatically display it with a slight dark-grey gradient; when the button is clicked, it will momentarily darken to solid black. This is an automatic "'button down' look" you didn't want; however, it is very subtle, and is a good indicator that the button worked. If you don't want this to happen, you could get an instance of the desired image and [stopImage setTemplate:NO];
on it.
Disable isBordered
let button = NSButton(
image: NSImage(named: NSImage.Name("plus"))!,
target: self,
action: #selector(onButtonPress)
)
button.isBordered = false
If you don't want to use a templated but want the push down highlight anyways, you can also use the following setup for an NSButton
:
let imageButton = NSButton()
imageButton.image = NSImage(named: "MyImage")!
imageButton.bezelStyle = .shadowlessSquare
imageButton.isBordered = false
imageButton.imagePosition = .imageOnly
The important thing to make the highlight work on any image is to set bezelStyle
to shadowlessSquare
.
I know this behavior wasn't requested in the question, but it might be useful for others.
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